Assume $G$ is a connected locally compact group and $M$ is a maximal compact subgroup of $G$. Is $M$ connected too?
Maximal Compact Subgroups – Are They Connected in Connected Groups?
gr.group-theorylie-groups
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I guess you are referring to the Tits Bourbaki seminar talk #119 here in the 1950s on subalgebras of semisimple Lie algebras (which translates the original question about compact Lie groups)? That's freely available online, but Dynkin's earlier papers probably aren't (though AMS published them long ago in translation volumes). Apparently Dynkin's tables contained some errors, as tables often do, but there is a useful AMS/IP collection here with corrections and commentaries which may be the best current source for the full story.
I suspect there has been too little incentive for anyone to rework and publish all of Dynkin's arguments and tables, but Gary Seitz (along with Martin Liebeck) did publish a number of long papers generalizing this work to the setting of semisimple algebraic groups (giving a lot of details in a modern style).
ADDED: There are two important (and long) papers by Dynkin in 1952, published by AMS in Series 2, Volume 6 of their translation series (1957), pages 111-378: Semisimple subalgebras of semisimple Lie algebras and Maximal subgroups of the classical groups. The papers contain lots of tables and are related to each other, as discussed by Tits in his Bourbaki talk. The main results are recovered by Seitz (and Liebeck) in their large AMS Memoir papers (such as No. 365 in 1987) treating maximal subgroups of semisimple algebraic groups. But this is a more elaborate framework, getting into prime characteristic as well.
To apply Dynkin's results to compact Lie groups one has to use Weyl's approach: complexify the Lie algebra or in reverse take a compact real form of a complex Lie algebra. Since a compact connected group is semisimple (or trivial) modulo its center and the latter lies in a torus, the semisimple case is crucial for maximal subgroup problems. The relevant maximal subalgebras of an associated complex Lie algebra are then semisimple. Unfortunately, compact groups don't seem to be treated explicitly in the literature (an exposition with some low rank groups as examples would be useful). But studying the group structure directly is too difficult, so the Lie algebra technology over $\mathbb{C}$ is most natural here, including some finite dimensional representation theory.
Surely it's easier to check whether $H'$ is simply-connected by inspecting the co-root lattice...? For the example of $G_2$ containing $SO(4)$ that Allen mentions, we have a pseudo-Levi subalgebra of type $A_1\times \tilde{A}_1$ where $\{(3\alpha+2\beta),\alpha\}$ is a basis of simple roots. Now the cocharacter $(3\alpha+2\beta)^\vee=\alpha^\vee+2\beta^\vee$, so the lattice:
${\mathbb Z}(3\alpha+2\beta)^\vee+{\mathbb Z}\alpha^\vee = {\mathbb Z}\alpha^\vee+{\mathbb Z}(2\beta^\vee)$
is of index two in the cocharacter lattice ${\mathbb Z}\alpha^\vee+{\mathbb Z}\beta^\vee$ for $T$.
In fact this allows you to determine exactly what the pseudo-Levi subgroup is in each case.
For the maximal pseudo-Levis there's an easier trick to find non-simply-connected ones: if $s\in T$ and $L=Z_G(s)$ then $Z(L)/Z(L)^\circ$ is generated by $s$, by a result of Eric Sommers. So we can see almost immediately that hardly any maximal pseudo-Levi subgroups are simply-connected. For example, the pseudo-Levi of $F_4$ which is of type $C_3\times A_1$ has a cyclic centre, so it can't be isomorphic to $Sp_6\times SL_2$. Specifically, it is isomorphic to $(Sp_6\times SL_2)/\{ \pm (I,I)\}$.
EDIT: A mistake with this is that Sommers' result only holds for adjoint type groups. More generally we have $Z(L)/(Z(L)^\circ Z(G))$ is generated by $s$. Of course this makes no difference for type $F_4$.
Best Answer
Disclaimer: Locally compact groups are absolutely not my field of expertise. I hope an expert can check my statements below, and perhaps add some details and references.
The Malcev–Iwasawa theorem implies that any connected, locally compact group $G$ satisfies:
$G$ has a maximal compact subgroup;
there exists $n\in\mathbb{N}$ such that for any maximal compact subgroup $K$ of $G$, the underlying space of $G$ is homeomorphic to $K\times\mathbb{R}^n$.
In particular, every maximal compact subgroup of a connected, locally compact group is itself connected.
References: The following references state the necessary results without proof.
Theorem 32.5 of Markus Stroppel's book "Locally compact groups".
The article "Compact subgroups of Lie groups and locally compact groups" (DOI: http://dx.doi.org/10.1090/S0002-9939-1994-1166357-9), published in the Proceedings of the American Mathematical Society, volume 120, number 2, in February 1994 (pages 623-634). See the statements of theorems A, B, and C in the introduction to this article. According to the discussion there, the theorems hold for connected, locally compact groups: they follow from the analogous results for Lie groups as soon as one knows that a connected, locally compact group is a projective limit of Lie groups.